In probability, an experiment is an activity whose outcome cannot be predicted with certainty in advance, though all possible outcomes are known. The collection of all such possible outcomes is called the sample space, usually denoted by \(S\).

An event is defined as any subset of the sample space.

In symbols, if Sis the sample space, then any set \[\boxed{\bbox[5pt]{E\subseteq S}}\] is called an event.

Thus, an event represents a specific situation or result we are interested in. It may consist of one outcome or several outcomes.

Example If a die is thrown once, \[S=\{1,2,3,4,5,6\}\] The event “getting an even number” is \[E=\{2,4,6\}\] which is clearly a subset of \(S\). Hence it is an event.

1. Simple (Elementary) Event:
   An event containing exactly one outcome is called a simple event.
   Example: In a die throw, the event \(\left\{3\right\}\).
2. Compound Event:
   An event containing more than one outcome is called a compound event. Example: \(\{2,\;4,\;6\}\).
3. Certain Event:
   The entire sample space itself is called the certain event, since it is sure to occur. \[E=S\]
4. Impossible Event:
   The empty set represents an impossible event and is denoted by \emptyset. \[E=\emptyset\]
5. Complementary Event:
   For any event \(E\), the event “not \(E\)” is called its complement and is denoted by \(E^\prime\). If \(E\subseteq S\), then \[E^\prime=S-E\]

Complementary Events

Let \(S\) be the sample space of a random experiment and let \(A\) be an event.
The event consisting of all outcomes of \(S\) which do not belong to \(A\) is called the complement of \(A\) and is denoted by \(A^\prime\).
Thus, \[A^\prime=S-A\] The pair \(A\) and \(A^\prime\) are called complementary events.

Meaning in simple words

• Event \(A\): “\(A\) happens.”
• Event \(A^\prime\): “\(A\) does not happen.”

Exactly one of these must occur in every trial.

Properties of Complementary Events

1. Mutually Exclusive

\[A\cap A^\prime=\emptyset\]

This means \(A\) and \(A^\prime\) cannot occur together.

2. Exhaustive

\[A\cup A^\prime=S\]

This means one of them must occur.

Probability Law for Complementary Events

Since \(A\) and \(A^\prime\) are mutually exclusive and exhaustive, \[P(A)+P(A^\prime)=P(S)=1\] Therefore, \[\fbox{$\bbox[5pt]{P(A^\prime)=1-P(A)}$}\] and also, \[\fbox{$\bbox[5pt]{P(A)=1-P(A^\prime)}$}\]

The Event ‘A or B’

Let \(A\) and \(B\) be two events related to the same random experiment.
The event “A or B” means that at least one of the two events occurs. It includes:

• outcomes where only Aoccurs,
• outcomes where only Boccurs,
• outcomes where both Aand Boccur.

Mathematically, this event is called the union of \(A\) and \(B\), and is written as \[A\mathrm{\ or\ }B=A\cup B\] So, \[A\cup B={\mathrm{all\ outcomes\ belonging\ to\ }A\mathrm{\ or\ }B\mathrm{\ or\ both}}\]

Probability of “\({A}\) or \({B}\)”

For any two events Aand B,

\[\fbox{$\bbox[5pt]{P(A\cup B)=P(A)+P(B)-P(A\cap B)}$}\]

The subtraction term removes the overlap (common outcomes counted twice).

Special Case (Mutually Exclusive Events)

If \(A\) and \(B\) cannot occur together, then

\[A\cap B=\emptyset\]

and the formula becomes

\[\boxed{\bbox[5pt]{P(A\cup B)=P(A)+P(B)}}\]

The Event ‘A and B’

Let Aand Bbe two events of the same random experiment.
The event “A and B” means that both events occur together.
Mathematically, this event is called the intersection of \(A\) and \(B\), and it is written as \[A\mathrm{\ and\ }B=A\cap B\] So, \[A\cap B={\mathrm{all\ outcomes\ common\ to\ both\ }A\mathrm{\ and\ }B}\]

Meaning in words

• Only those outcomes are included which belong to both Aand B.
• If there is no common outcome, then \(A\cap B=\emptyset\) (they cannot happen together).

Probability of “\(A\) and \(B\)”

The probability that both Aand Boccur is written as \[P(A\cap B)\] This value represents the likelihood of the common part of the two events.

The Event ‘A but not B’

Let \(A\) and \(B\) be two events of the same random experiment.
The event \(A\) but not \(B\) means that event \(A\) occurs while event \(B\) does not occur.
Mathematically, this event is written as \[A\mathrm{\ but\ not\ }B=A\cap B^\prime\] It is also called the difference of events and is sometimes written as \(A-B\).

So, \[A\cap B^\prime={\mathrm{all\ outcomes\ which\ belong\ to\ }A\mathrm{\ and\ not\ to\ }B}\]

Probability of \(A\) but not \(B\)

Since \[A=(A\cap B)\cup(A\cap B^\prime)\] and these two parts do not overlap, we get \[P(A)=P(A\cap B)+P(A\cap B^\prime)\] Rearranging, \[\fbox{$\bbox[5pt]{P(A\mathrm{\ but\ not\ }B)=P(A\cap B^\prime)=P(A)-P(A\cap B)}$}\] This formula is very useful in numerical problems.

Mutually exclusive events

Let \(A\) and \(B\) be two events of the same random experiment.
Two events are called mutually exclusive if they cannot occur at the same time.
In other words, there is no common outcome between them.

Mathematically,

\[A\cap B=\emptyset\] This means the intersection of Aand Bis the empty set.

Meaning in simple words

• If \(A\) happens, \(B\) cannot happen.
• If \(B\) happens, \(A\) cannot happen.
• They are completely separate possibilities.

Probability Result for Mutually Exclusive Events

Since there is no overlap,

\[n(A\cup B)=n(A)+n(B)\]

Dividing by the total number of outcomes \(n(S)\),

\[\fbox{$\bbox[5pt]{P(A\cup B)=P(A)+P(B)}$}\]

So, for mutually exclusive events, the probability of “A or B” is simply the sum of their individual probabilities.

Let \(S\) be the sample space of a random experiment and let \(A,\;B,\;C,\;\ldots\) be events associated with it.
A set of events is called exhaustive if together they cover the entire sample space.
That is, at least one of these events must occur whenever the experiment is performed.
Mathematically, events Aand Bare exhaustive if

\[A\cup B=S\]

For three events \(A,\;B,\;C\),

\[A\cup B\cup C=S\]

Meaning in simple words

• Exhaustive events leave no possible outcome uncovered.
• Every result of the experiment belongs to at least one of the given events.

Probability Result for Exhaustive Events

If \(A\) and \(B\) are exhaustive, then \[A\cup B=S\]

Taking probabilities,

\[P(A\cup B)=P(S)=1\]

Using the addition formula,

\[P(A)+P(B)-P(A\cap B)=1\]

This relation is often used in numerical problems.

Special Case:

Exhaustive and Mutually Exclusive
If two events are both exhaustive and mutually exclusive, then \[A\cup B=S\mathrm{and}A\cap B=\emptyset\] So, \[\fbox{$\bbox[5pt]{P(A)+P(B)=1}$}\]

Axiomatic Approach to Probability

The axiomatic approach is a formal and systematic way of defining probability. Instead of relying on intuition, this method is based on a few basic rules, called axioms, which are used to assign probabilities to events.

Let \( S \) be the sample space of a random experiment. The probability function \( P \) assigns a real number to each event in the power set of \( S \). The value of \( P \) always lies between 0 and 1, and it satisfies the following axioms.

For any event \( E \), the probability is non-negative, that is, \( P(E) \geq 0 \).

The probability of the sample space is 1, so \( P(S) = 1 \).

If two events \( E \) and \( F \) are mutually exclusive, then the probability of their union is equal to the sum of their probabilities, that is, \( P(E \cup F) = P(E) + P(F) \).

From this third axiom, we can show that the probability of the impossible event is zero. Let \( \varnothing \) denote the empty set. Since any event \( E \) and \( \varnothing \) are disjoint, we have

\[ P(E \cup \varnothing) = P(E) + P(\varnothing) \]

But \( E \cup \varnothing = E \), therefore

\[ P(E) = P(E) + P(\varnothing) \]

This is possible only if \( P(\varnothing) = 0 \).

Now consider a sample space

\[ S = \{\omega_1, \omega_2, \ldots, \omega_n\} \]

where each \( \omega_i \) represents an outcome of the experiment. From the axiomatic definition of probability, it follows that

\[ 0 \leq P(\omega_i) \leq 1 \quad \text{for each } \omega_i \in S \]

The sum of the probabilities of all outcomes is equal to 1, that is,

\[ P(\omega_1) + P(\omega_2) + \cdots + P(\omega_n) = 1 \]

For any event \( A \), the probability of \( A \) is obtained by adding the probabilities of all outcomes that belong to \( A \). Hence,

\[ P(A) = \sum P(\omega_i), \quad \omega_i \in A \]

In everyday life, we often come across situations where the outcome of an action is uncertain. For example, when a coin is tossed, we cannot say for sure whether the result will be a head or a tail. The mathematical measure used to express the likelihood of such outcomes is called probability.

An experiment is a process that leads to one of several possible results, known as outcomes. The collection of all possible outcomes of an experiment is called the sample space, usually denoted by S. Any subset of the sample space is called an event. If an event consists of only one outcome, it is called a simple (or elementary) event.

When all outcomes of an experiment are equally likely, the probability of an event Eis defined as the ratio of the number of outcomes favourable to Eto the total number of possible outcomes in the sample space.
Mathematically, the probability of an event Eis given by
\[P(E)=\frac{\mathrm{Number\ of\ favourable\ outcomes\ to\ }E}{\mathrm{Total\ number\ of\ outcomes\ in\ }S}\]

The value of probability always lies between 0 and 1, inclusive. If \(P(E)=0\), the event is impossible, and if \(P(E)=1\), the event is certain to occur. An event that cannot occur is called an impossible event, while an event that must occur is called a sure event.

If an experiment has a finite number of equally likely outcomes, then the probability of any event can be calculated easily using the above formula. Thus, probability provides a clear and logical way to study uncertainty and make predictions based on mathematical reasoning.

Consider the experiment of rolling a die. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Write the sets representing the events
(i) A or B
(ii) A and B
(iii) A but not B
(iv) ‘not A’.

Solution

For one roll of a die, the sample space is

$$ \begin{aligned} S&=\{1,2,3,4,5,6\} \\ A&=\{2,3,5\} \\ B&=\{1,3,5\} \end{aligned} $$

Here, \(A\) represents the prime numbers on a die and \(B\) represents the odd numbers.

For part (i), “\(A\) or \(B\)” means all outcomes which belong to \(A\) or \(B\) or both. Hence,

$$ A\cup B=\{1,2,3,5\} $$

For part (ii), “\(A\) and \(B\)” means the common outcomes of \(A\) and \(B\). Therefore,

$$ A\cap B=\{3,5\} $$

For part (iii), “\(A\) but not \(B\)” means those elements which belong to \(A\) but do not belong to \(B\). Thus,

$$ A-B=\{2\} $$

For part (iv), “not \(A\)” represents the complement of \(A\), that is, all elements of the sample space which are not in \(A\). Hence,

$$ A'=\{1,4,6\} $$

These sets are obtained directly from the definitions of union, intersection, difference, and complement of events.

Two dice are thrown and the sum of the numbers which come up on the dice is noted. Let us consider the following events associated with this experiment
A: ‘the sum is even’.
B: ‘the sum is a multiple of 3’.
C: ‘the sum is less than 4’.
D: ‘the sum is greater than 11’.
Which pairs of these events are mutually exclusive?

Solution

When two dice are thrown, each outcome is an ordered pair \((i,j)\), where \(i,j\in\{1,2,3,4,5,6\}\).

$$ \begin{aligned} A&=\{(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),\\ &\quad(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)\} \\ B&=\{(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3),(6,6)\} \\ C&=\{(1,1),(1,2),(2,1)\} \\ D&=\{(6,6)\} \end{aligned} $$

Here, \(A\) consists of all outcomes for which the sum is even, \(B\) consists of all outcomes for which the sum is a multiple of 3, \(C\) contains outcomes for which the sum is less than 4, and \(D\) contains outcomes for which the sum is greater than 11.

Now, event \(C\) corresponds to sums 2 and 3, while event \(D\) corresponds only to sum 12. Clearly, no ordered pair can satisfy both conditions simultaneously.

$$ C\cap D=\varnothing $$

Hence, \(C\) and \(D\) have no common outcomes and therefore form a pair of mutually exclusive events. All other pairs have at least one common outcome, so they are not mutually exclusive.

A coin is tossed three times, consider the following events.
A: ‘No head appears’,
B: ‘Exactly one head appears’ and
C: ‘Atleast two heads appear’.
Do they form a set of mutually exclusive and exhaustive events?

Solution

When a coin is tossed three times, the sample space is

$$ \begin{aligned} S&=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\} \\ A&=\{TTT\} \\ B&=\{HTT,THT,TTH\} \\ C&=\{HHT,HTH,THH,HHH\} \end{aligned} $$

Here, \(A\) represents the event that no head appears, \(B\) represents the event that exactly one head appears, and \(C\) represents the event that at least two heads appear.

$$ \begin{aligned} A\cap B=\varnothing \\ A\cap C=\varnothing \\ B\cap C=\varnothing \end{aligned} $$

Since no outcome is common to any two of the events, they are mutually exclusive.

$$ A\cup B\cup C=S $$

Further, every possible outcome of the experiment belongs to exactly one of the events \(A\), \(B\), or \(C\). Hence, together they cover the entire sample space.

Therefore, \(A\), \(B\), and \(C\) form a set of mutually exclusive and exhaustive events.

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be
(i) a diamond
(ii) not an ace
(iii) a black card (i.e., a club or, a spade)
(iv) not a diamond
(v) not a black card.

Solution

One card is drawn at random from a well-shuffled deck of 52 cards. Since all outcomes are equally likely, the probability of any event is given by the ratio of favourable outcomes to total outcomes.

(i) Probability that the card drawn is a diamond. There are 13 diamond cards in the deck.

$$ \begin{aligned} P(E)&=\dfrac{13}{52} \\ &=\dfrac{1}{4} \end{aligned} $$

(ii) Probability that the card drawn is not an ace. First, the probability of drawing an ace is calculated. There are 4 aces in the deck.

$$ \begin{aligned} P(E)&=\dfrac{4}{52} \\ &=\dfrac{1}{13} \\\\ P'(E)&=1-\dfrac{1}{13} \\ &=\dfrac{12}{13} \end{aligned} $$

Hence, the probability that the card drawn is not an ace is \(\dfrac{12}{13}\).

(iii) Probability that the card drawn is black. There are 26 black cards (clubs and spades) in the deck.

$$ \begin{aligned} P(E)&=\dfrac{26}{52} \\ &=\dfrac{1}{2} \end{aligned} $$

(iv) Probability that the card drawn is not a diamond. Using the complement rule and the result from part (i),

$$ \begin{aligned} P(E)&=\dfrac{13}{52} \\ &=\dfrac{1}{4} \\ P'(E)&=1-\dfrac{1}{4} \\ &=\dfrac{3}{4} \end{aligned} $$

(v) Probability that the card drawn is not black. Since the probability of drawing a black card is \(\dfrac{1}{2}\),

$$ \begin{aligned} P(E)&=\dfrac{26}{52} \\ &=\dfrac{1}{2} \\ P'(E)&=1-\dfrac{1}{2} \\ &=\dfrac{1}{2} \end{aligned} $$

Thus, the required probabilities are obtained using direct counting and the complement principle.

A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be
(i) red,
(ii) yellow,
(iii) blue,
(iv) not blue,
(v) either red or blue.

Solution

The bag contains 9 discs in total, of which 4 are red, 3 are blue, and 2 are yellow. Since all discs are similar in shape and size, each disc is equally likely to be drawn.

Red discs = 4, Blue discs = 3, Yellow discs = 2

(i) Probability that a red disc is drawn

$$ P(R)=\dfrac{4}{9} $$

(ii) Probability that a yellow disc is drawn

$$ P(Y)=\dfrac{2}{9} $$

(iii) Probability that a blue disc is drawn. Since there are 3 blue discs out of 9,

$$ \begin{aligned} P(B)&=\dfrac{3}{9} \\ &=\dfrac{1}{3} \end{aligned} $$

(iv) Probability that the disc drawn is not blue. Using the complement rule,

$$ \begin{aligned} P(B)&=\dfrac{1}{3} \\ P'(B)&=1-\dfrac{1}{3} \\ &=\dfrac{2}{3} \end{aligned} $$

Hence, the probability that a disc other than blue is drawn is \(\dfrac{2}{3}\).

(v) Probability that either a red or a blue disc is drawn. Since a disc cannot be both red and blue at the same time, these events are mutually exclusive.

$$ \begin{aligned} P(R\text{ or }B)&=P(R)+P(B) \\ &=\dfrac{4}{9}+\dfrac{1}{3} \\ &=\dfrac{4}{9}+\dfrac{3}{9} \\ &=\dfrac{7}{9} \end{aligned} $$

Thus, the required probabilities are obtained directly by counting favourable outcomes and using the complement principle where needed.

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that
(a) Both Anil and Ashima will not qualify the examination.
(b) Atleast one of them will not qualify the examination and
(c) Only one of them will qualify the examination.

Solution

Let \(E\) denote the event that Anil qualifies the examination and \(F\) denote the event that Ashima qualifies the examination. The given data are

$$ \begin{aligned} P(E)&=0.05,\\ P(F)&=0.10,\\ P(E\cap F)&=0.02 \end{aligned} $$

(a) Probability that both Anil and Ashima will not qualify

$$ \begin{aligned} E'\cap F'&=(E\cup F)' \\ P(E\cup F)&=P(E)+P(F)-P(E\cap F) \\ &=0.05+0.10-0.02 \\ & =0.13 \\\\ P(E'\cap F')&=1-P(E\cup F) \\ &=1-0.13 \\ &=0.87 \end{aligned} $$

(b) Probability that at least one of them will not qualify. This is the complement of the event that both qualify.

$$ \begin{aligned} 1-P(E\cap F) &=1-0.02 \\ &=0.98 \end{aligned} $$

(c) Probability that only one of them will qualify, that is, either Anil qualifies and Ashima does not, or Ashima qualifies and Anil does not

$$ \begin{aligned} \big(E\cap F'\big)\cup\big(E'\cap F\big) &=P(E\cap F')+P(E'\cap F) \\ &=\big[P(E)-P(E\cap F)\big]+\big[P(F)-P(E\cap F)\big] \\ &=0.05-0.02+0.10-0.02 \\ &=0.11 \end{aligned} $$

Thus, the required probabilities are obtained by using the addition rule and the complement principle.

A committee of two persons is selected from two men and two women. What is the probability that the committee will have
(a) no man?
(b) one man?
(c) two men?

Solution

There are 2 men and 2 women, so a total of 4 persons. A committee of 2 persons is to be formed.

The total number of possible committees is

$$ \begin{aligned} {}^{4}C_{2}&=\dfrac{4\times3\times2!}{2!(4-2)!} \\ &=\dfrac{4\times3}{2\times1} \\ &=6 \end{aligned} $$

(a) Probability that no man is selected, which means both members are women

$$ \begin{aligned} P(E)&=\dfrac{{}^{2}C_{2}}{{}^{4}C_{2}} \\ &=\dfrac{1}{6} \end{aligned} $$

(b) Probability that exactly one man is selected. This can happen by choosing 1 man from 2 men and 1 woman from 2 women.

$$ \begin{aligned} {}^{2}C_{1}\times{}^{2}C_{1} &=2\times2 \\ &=4 \\\\ P(E)&=\dfrac{4}{6} \\ &=\dfrac{2}{3} \end{aligned} $$

(c) Probability that two men are selected

$$ \begin{aligned} P(E)&=\dfrac{{}^{2}C_{2}}{{}^{4}C_{2}} \\ &=\dfrac{1}{6} \end{aligned} $$

Thus, the required probabilities are obtained by counting favourable committees in each case and dividing by the total number of possible committees.